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The molecules \(\mathrm{SiF}_{4}, \mathrm{SF}_{4},\) and \(\mathrm{XeF}_{4}\) have molecular formulas of the type \(\mathrm{AF}_{4}\), but the molecules have different molecular geometries. Predict the shape of each molecule, and explain why the shapes differ.

Short Answer

Expert verified
SiF4 has a tetrahedral geometry with four bonding electron domains around the Si atom. SF4 has a see-saw geometry with five electron domains (four bonding and one lone pair) around the S atom. XeF4 has a square planar geometry with six electron domains (four bonding and two lone pairs) around the Xe atom. The difference in molecular geometries is due to the differences in the number and arrangement of bonding and non-bonding electron domains around their central atoms, as explained by the VSEPR theory.

Step by step solution

01

Determine the electron domains around the central atom

In SiF4, Si is the central atom, and each of the four F atoms is bonded to it. Thus, there are four bonding electron domains around Si. In SF4, S is the central atom, with four F atoms bonded to it and one lone pair of electrons. Therefore, there are five electron domains around S - four bonding domains and one non-bonding (lone pair) domain. In XeF4, Xe is the central atom, with four F atoms bonded to it and two lone pairs of electrons. Thus, there are six electron domains around Xe - four bonding domains and two non-bonding (lone pair) domains.
02

Use VSEPR theory to determine molecular geometries

For SiF4, there are four bonding electron domains, which means the electron geometry is tetrahedral. Since there are no lone pairs, the molecular geometry is also tetrahedral. For SF4, there are five electron domains, resulting in a trigonal bipyramidal electron geometry. However, since there is one lone pair, the molecular geometry is called "see-saw" or distorted tetrahedron. For XeF4, there are six electron domains, thus the electron geometry is octahedral. With two lone pairs, the molecular geometry is square planar.
03

Explain the difference in molecular shapes

The molecules SiF4, SF4, and XeF4 have different molecular geometries due to the differences in the number of electron domains (bonding and non-bonding) around their central atoms. In SiF4, the absence of lone pairs leads to a tetrahedral geometry. For SF4, the presence of one lone pair results in a see-saw geometry. In XeF4, the presence of two lone pairs leads to a square planar geometry. The VSEPR theory helps in understanding the molecular shapes, as it suggests that the electron domains will arrange themselves to minimize electron pair repulsion. This leads to distinct molecular geometries based on the number and arrangement of bonding and non-bonding electron domains.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

VSEPR Theory
VSEPR theory, or Valence Shell Electron Pair Repulsion theory, is a cornerstone in the study of molecular geometry. It offers a straightforward method for predicting the shape of individual molecules based on the number of electron pairs surrounding their central atoms.

The foundation of the VSEPR theory lies in the concept that electron pairs within a molecule repel each other as they contain negatively charged electrons. This repulsion force influences the spatial arrangement of the atoms within a molecule, which ultimately gives the molecule its shape. Both bonding electron pairs (which are shared between atoms in a chemical bond) and lone pairs (which are not shared, but reside solely on the central atom) have a significant impact.

In essence, molecules organize themselves to minimize the repulsive forces between electron pairs, leading to distinct geometries. This theory is especially useful because it provides a model to predict molecular structure that can be validated with real-world observations and experiments.
Electron Domains
Electron domains are regions around a molecule's central atom where electrons are most likely to be found. These domains are categorized into two types: bonding domains, which are also known as bonding pairs, meaning they are shared by two atoms; and non-bonding domains, often referred to as lone pairs, which are localized on a single atom.

To determine the molecular geometry of a molecule, it's essential to count all electron domains. Bonding domains contribute to the shape of the molecule, while non-bonding domains exert a greater repulsive force, differing in molecular geometry compared to the electron-domain geometry.

Understanding Electron Domains

For instance, the number of electron domains dictates the base electron geometry, with arrangements such as linear, trigonal planar, tetrahedral, trigonal bipyramidal, and octahedral. These geometries correspond to 2, 3, 4, 5, and 6 electron domains, respectively. It is crucial to remember that the VSEPR theory considers both types of electron domains to predict molecular shapes effectively.
Molecular Shapes
Molecular shapes are the three-dimensional structures attributed to molecules, which are determined by the positions of the nuclei and the electron domains. Understanding molecular shapes is critical for grasping the nature of chemical reactions, physical properties of substances, and biological interactions.

The VSEPR theory allows us to predict these shapes by considering the repulsion between electron domains. As the number of electron domains increases, the complexity of the possible molecular shapes also increases.

Real-world Examples

SiF4, with four bonding domains and no lone pairs, forms a tetrahedral shape. SF4, with one lone pair and four bonding domains, adopts a see-saw shape. This deviation from the base trigonal bipyramidal structure occurs because the lone pair occupies more space, leading to a distortion. Lastly, XeF4 demonstrates that with two lone pairs amidst six electron domains, a square planar geometry is favored. Each example showcases how variations in bonding and lone pair arrangements alter molecular structures, resulting in diverse geometries even among molecules with the same type and number of atoms.

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Most popular questions from this chapter

Predict whether each of the following molecules is polar or nonpolar: (a) IF, (b) \(\mathrm{CS}_{2}\) (c) \(\mathrm{SO}_{3}\) (d) \(\mathrm{PCl}_{3}\) (e) \(\mathrm{SF}_{6}\) (f) \(\mathrm{IF}_{5}\)

(a) The nitric oxide molecule, NO, readily loses one electron to form the \(\mathrm{NO}^{+}\) ion. Why is this consistent with the electronic structure of \(\mathrm{NO} ?\) (b) Predict the order of the \(\mathrm{N}-\mathrm{O}\) bond strengths in \(\mathrm{NO}, \mathrm{NO}^{+},\) and \(\mathrm{NO}^{-},\) and describe the magnetic properties of each. (c) With what neutral homonuclear diatomic molecules are the \(\mathrm{NO}^{+}\) and \(\mathrm{NO}^{-}\) ions isoelectronic (same number of electrons)?

Propylene, \(\mathrm{C}_{3} \mathrm{H}_{6},\) is a gas that is used to form the important polymer called polypropylene. Its Lewis structure is (a) What is the total number of valence electrons in the propylene molecule? (b) How many valence electrons are used to make \(\sigma\) bonds in the molecule? (c) How many valence electrons are used to make \(\pi\) bonds in the molecule? (d) How many valence electrons remain in nonbonding pairs in the molecule? (e) What is the hybridization at each carbon atom in the molecule?

(a) Draw Lewis structures for ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\), ethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\), and acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right) .\) (b) What is the hybridization of the carbon atoms in each molecule? (c) Predict which molecules, if any, are planar. (d) How many \(\sigma\) and \(\pi\) bonds are there in each molecule? (e) Suppose that silicon could form molecules that are precisely the analogs of ethane, ethylene, and acetylene. How would you describe the bonding about Si in terms of hydrid orbitals? Silicon does not readily form some of the analogous compounds containing \(\pi\) bonds. Why might this be the case?

In which of the following molecules can you confidently predict the bond angles about the central atom, and for which would you be a bit uncertain? Explain in each case. (a) \(\mathrm{H}_{2} \mathrm{~S}\), (b) \(\mathrm{BCl}_{3}3,\) (c) \(\mathrm{CH}_{3} \mathrm{I},(\mathrm{d}) \mathrm{CBr}_{4}\), (e) \(\mathrm{TeBr}_{4}\)

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